Question

Consider the following multistep reaction: A+B→AB(slow) A+AB→A2B(fast)–––––––––––––––––––– 2A+B→A2B(overall) Based on this mechanism, determine the rate law for the overall reaction. Express your answer in standard MasteringChemistry format. For example, if the rate law is k[A]3[B]2 type k*[A]^3*[B]^2. View Available Hint(s)

Answer 1

Answer : The rate law for the overall reaction is, $Rate=k[A][B]$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

As we are given the mechanism for the reaction :

Step 1 : $A+B\rightleftharpoons AB$    (slow)

Step 2 : $A+AB\rightarrow A_2B$    (fast)

$2A+B\rightarrow A_2B$      (overall)

The rate law expression for overall reaction should be in terms of A and B.

As we know that the slow step is the rate determining step. So,

The slow step reaction is,

$A+B\rightleftharpoons AB$

The expression of rate law for this reaction will be,

$Rate=k[A][B]$

Hence, the rate law for the overall reaction is, $Rate=k[A][B]$